SOLUTION: Find the equation of a st.line through (-3,2) that is perpendicular to x+3y=6. Write the equation of the st.line in standard form. *So far I have X+3y=6 -X -X

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Find the equation of a st.line through (-3,2) that is perpendicular to x+3y=6. Write the equation of the st.line in standard form. *So far I have X+3y=6 -X -X       Log On


   

Question 204733: Find the equation of a st.line through (-3,2) that is perpendicular to x+3y=6. Write the equation of the st.line in standard form.
*So far I have X+3y=6
-X -X
=3y= -x+6 but according to my notes its im missing a step.
Can you please help.
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
perpendicular lines have slopes that are negative reciprocals of each other

your next step is to get y by itself by dividing by 3 ___ y = -1/3 x + 2 ___ this is the slope-intercept form
___ the slope (m) is the coefficient of the x term ___ y = m x + b
___ m = -1/3

so the slope of the new line is 3 (negative reciprocal of -1/3)

using point-slope ___ y - 2 = 3 (x - (-3)) ___ y = 3x + 11